Download - Lecture 24

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Page 1: Lecture 24

Lecture – 24

Calculus of Variations : An Overview

Dr. Radhakant PadhiAsst. Professor

Dept. of Aerospace EngineeringIndian Institute of Science - Bangalore

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ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

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Fundamental Theorems of Calculus

Theorem – 1:

Theorem – 2:

Theorem – 3:

( ) ( ) ( ), provided is continuousx

a

d f d f x f xdx

σ σ =∫

( ) ( )

( ) ( )

,,

provided , has continuous partial derivative /

b b

a a

f x yd f x y dy dydx x

f x y f x

∂=

∂ ∂

∫ ∫

( ) ( ) ( ) ( )2 2

1 1

( ) ( )2 1

2 1( ) ( )

1 2

,, , ( ) , ( )

provided , , have continuous partial derivatives with respect to

x x

x x

f x y d dd f x y dy dy f x x f x xdx x dx dx

f x

ψ ψ

ψ ψ

ψ ψψ ψ

ψ ψ

∂ ⎡ ⎤= + −⎢ ⎥∂ ⎣ ⎦∫ ∫

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ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

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Calculus of Variations:Basic Concepts

Function (to each value of the independent variable, there is a corresponding value of the dependent variable)

Increment of a function

Functional (to each function, there is a corresponding value of the dependent variable)

Increment of a functional

( ) 32 3x t t t= + ( )( ) ( )

( )

1

01

3

0

2 3 2

J x t x t dt

t t dt

=

= + =

( ) ( )x x t t x tΔ + Δ − ( ) ( )( ) ( )( )J J x t x t J x tδΔ + −

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ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

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Example

( ) ( )

( )

( )( )

( )( )( )( )

0 0

0

0

0

2 2

2 2

22 2

2

( ) ( ) ( )

2 ( ) ( ) 1 2 ( ) 1

2 ( ) ( ) 1 2 ( ) 1

2 ( ) 2 ( ) ( ) ( ) 1 2 ( ) 1

4 ( ) ( ) 2 ( )

f f

f

f

f

t t

t t

t

t

t

t

t

t

J J x t x t J x t

x t x t dt x t dt

x t x t x t dt

x t x t x t x t x t dt

x t x t x t dt

δ

δ

δ

δ δ

δ δ

Δ = + −

⎡ ⎤ ⎡ ⎤= + + − +⎣ ⎦⎣ ⎦

⎡ ⎤ ⎡ ⎤= + + − +⎣ ⎦⎣ ⎦

⎡ ⎤ ⎡ ⎤= + + + − +⎣ ⎦⎣ ⎦

⎡ ⎤= +⎣ ⎦

∫ ∫

0

22 ( ) 1ft

t

J x t dt⎡ ⎤= +⎣ ⎦∫

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ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

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Calculus of Variations:Basic Concepts

Function and its increment Functional and its increment

Reference: D. S. Naidu, Optimal Control Systems, CRC Press, 2002.

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ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

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Calculus of Variations:Basic Concepts

Differential of a function Variation of a functional

( ) ( )

( )

( )

* *

2

*

* * *

22

2

: First deviation : Second deviation

2

* *

0 0

12!

lim lim

. . in general

t t

df d f

t tt

f f t t f t

df d ft tdt dt

df d f

dfdf f tdt

i e df f t

Δ → Δ →

Δ = + Δ −

⎛ ⎞⎛ ⎞= Δ + Δ +⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= + +

⎛ ⎞= Δ = Δ⎜ ⎟

⎝ ⎠

= Δ

( ) ( )

( )2

22

2

: First variation : Second variation

2

( ) ( ) ( )

12!

J J

J J x t x t J x t

J Jx xx x

J J

JJ xx

δ δ

δ

δ δ

δ δ

δ δ

Δ = + −

⎛ ⎞∂ ∂⎛ ⎞= + +⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

= + +

∂⎛ ⎞= ⎜ ⎟∂⎝ ⎠

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ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

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Calculus of Variations:Basic Concepts

Result – 1:Derivative of variation = Variation of derivative

Result – 2:Integration of variation = Variation of integration

( ) ( ) ( )

( ) ( )

( )

*

*

d dx t x t x tdt dt

dx t dx tdt dt

x t

δ

δ

⎡ ⎤⎡ ⎤ = −⎣ ⎦ ⎣ ⎦

= −

⎡ ⎤= ⎣ ⎦

( ) ( ) ( )

( ) ( )

( )

0 0

0 0

0

*

*

t t

t t

t t

t t

t

t

x d x x d

x d x d

x d

δ τ τ τ τ τ

τ τ τ τ

δ τ τ

⎡ ⎤= −⎣ ⎦

= −

⎡ ⎤= ⎢ ⎥

⎢ ⎥⎣ ⎦

∫ ∫

∫ ∫

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ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

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Exercise

( )0

2

Evaluate the variation of :

( ) 2 ( ) 3 ( ) 4

Note: By 'variation', we mean 'first variation' (by default)

ft

t

J x t x t x t dt⎡ ⎤= + +⎣ ⎦∫

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ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

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Solution

( ) ( )

[ ]

0 0

0

0

0

2 2

2 2 2

2

:( ) ( ) ( )

2( ( ) ( )) 3( ( ) ( )) 4 2 ( ) 3 ( ) 4

2 2 ( ) 3( ) 4 2 3 4

4 2( ) 3

4 3 Neglect

f f

f

f

f

t t

t t

t

t

t

t

t

t

J J x t x t J x t

x t x t x t x t dt x t x t dt

x x x x x x x x dt

x x x x dt

x xdt

δ

δ δ

δ δ δ

δ δ δ

δ

Δ = + −

⎡ ⎤ ⎡ ⎤= + + + + − + +⎣ ⎦ ⎣ ⎦

⎡ ⎤⎡ ⎤ ⎡ ⎤= + + + + + − + +⎣ ⎦ ⎣ ⎦⎣ ⎦

⎡ ⎤= + +⎣ ⎦

= +

∫ ∫

Method - 1

( )ing the higher order term

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ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

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[ ]

0

0

0

2

2

2 ( ) 3 ( ) 4 ( )

2 3 4

(Variation of integral = Integral of variation)

4 3

f

f

f

t

t

t

t

t

t

JJ xx

x t x t dt x tx

x x x dtx

x xdt

δ δ

δ

δ

δ

∂⎛ ⎞= ⎜ ⎟∂⎝ ⎠⎡ ⎤∂ ⎡ ⎤= + +⎢ ⎥⎣ ⎦∂ ⎢ ⎥⎣ ⎦

⎡ ⎤∂ ⎡ ⎤= + +⎢ ⎥⎣ ⎦∂⎢ ⎥⎣ ⎦

= +

Method - 2 :

Solution

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ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

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Boundary Conditions

Fixed End Point Problems

Free End Point Problems• Completely free

• May be required to lie on a curve

( )( )

0 0, ( ) : Specified

, ( ) : Specifiedf f

t x t

t x tt

( )x t

( )0 0, ( )t x t

( ), ( )f ft x t

( )tη

t

( )x t

( )0 0, ( )t x t

( )tη

( ), ( )f ft x t

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ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

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Optimum of a Functional( )

( ) ( ) ( )

( ) ( ) ( )

*

*

* *

A functional is said to have a relative optimum at , if 0

such that for all functions which satisfy ,

the increment of has the "same sign".

1) If 0, then is a r

x t

x t x t x t

J

J J x J x J x

ε

ε

∃ >

∈Ω − <

Δ = − ≥

( ) ( ) ( )

( )

* *

*

elative (local) "Minimum".

2) If 0, then is a relative (local) "Maximum".

Note: If the above relationships are satisfied for arbitrarily large 0,

then is a "global optimum".

J J x J x J x

J x

ε

Δ = − ≤

>

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ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

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Optimum of a Functional

a b t

( )* : Optimum pathx t

( )x t

( ) ( ) ( )*Variation: x t x t x tδ = −*,x x

( )x tδ

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ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

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Fundamental Theorem of Calculus of Variations

( )

( )

*

*

For to be a candidate for optimum, the following conditions hold good:

1) Necessary Condition: ( ), ( ) 0, admissible ( )

2) Sufficiency Condition:

x t

J x t x t x tδ δ δ= ∀

2

2

0 (for minimum) 0 (for maximum)

JJ

δ

δ

>

<

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ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

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Fundamental Lemma

( )

( ) ( )

( )

( )

f

0

t

t

0

0

If for every continuous function

0

where the variation is continuous in , ,

then

0 ,

f

f

g t

g t x t dt

x t t t t

g t t t t

δ

δ

=

⎡ ⎤∈ ⎣ ⎦

⎡ ⎤= ∀ ∈⎣ ⎦

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ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

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Necessary condition of optimalityProblem: ( ) ( ) ( )

0

0

Optimize , , by appropriate selection of .

Note: , are fixed.

ft

t

f

J L x t x t t dt x t

t t

⎡ ⎤= ⎣ ⎦∫

Solution: ( )Make sure 0 for arbitrary J x tδ δ=

1) Euler – Lagrange (E-L) Equation

2) Transversality (Boundary) Condition

0L d Lx dt x∂ ∂⎛ ⎞− =⎜ ⎟∂ ∂⎝ ⎠

0

0ft

t

L xxδ∂⎡ ⎤ =⎢ ⎥∂⎣ ⎦

Note: Part of this equation mayalready be satisfied by problem formulation

Necessary Conditions:

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ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

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Proof:Hint: Use the necessary condition of optimality from the Fundamental theorem

( )( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ){ }0 0

0 0

*0

*

* * *

* *

Let , [ , ] : Optimumsolution

( ) : Some adjacent solution

Then

, , , ,

, , , ,

f f

f f

f

t t

t t

t t

t tL

x t t t t

x t x t

J J J L x t x t t dt L x t x t t dt

L x t x t t L x t x t t dt L dt

δ

Δ

+

⎡ ⎤⎡ ⎤Δ = − = −⎣ ⎦ ⎣ ⎦

⎡ ⎤⎡ ⎤= − = Δ⎣ ⎦ ⎣ ⎦

∫ ∫

∫ ∫

a b t

( )* : Optimum pathx t

( )x t

( ) ( ) ( )*Variation: x t x t x tδ = −*,x x

( )x tδ

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ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

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( ) ( )

* * * *

* * * *

, ,

However at every point ,

, , , ,

Assumption: is continuous and smooth in both and .

Then, in the limit,

f fx x x x

t

L L x x x x t L x t x t t

L Lx x HOTx x

L x xL LL L xx x

δ δ

δ δ

δ δ

⎡ ⎤ ⎡ ⎤Δ = + + −⎣ ⎦ ⎣ ⎦∂ ∂

= + +∂ ∂

∂ ∂Δ → = +

∂ ∂. In that case,xδ⎡ ⎤

⎢ ⎥⎣ ⎦

Proof

0

ft

t

L LJ J x x dtx x

δ δ δ∂ ∂⎡ ⎤Δ → = +⎢ ⎥∂ ∂⎣ ⎦∫

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ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

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0 0

0 0

0

( )

( ) ( )

f f

ff f

o

ff

o

t t

t t

tt t

t tt

tt

tt

L L d xx dt dtx x dt

L d x d L d xdt dt dtx dt dt x dt

L d Lx xdtx dt x

δδ

δ δ

δ δ

∂ ⎡ ∂ ⎤⎡ ⎤ ⎛ ⎞= ⎜ ⎟⎢ ⎥⎢ ⎥∂ ∂⎣ ⎦ ⎝ ⎠⎣ ⎦

⎡ ⎤∂ ⎡ ∂ ⎤⎛ ⎞ ⎛ ⎞ ⎡ ⎤= −⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥∂ ∂⎝ ⎠ ⎝ ⎠ ⎣ ⎦⎣ ⎦⎢ ⎥⎣ ⎦

⎡ ∂ ⎤ ⎡ ∂ ⎤⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥∂ ∂⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

∫ ∫

∫ ∫ ∫

ProofHowever,

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ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

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Proof

0

0 0

0

0 (Necessary condition of optimali

f

f ff

o

f f

o

t

t

t tt

t tt

t t

t t

L LJ x x dtx x

L L d Lxdt x xdtx x dt x

L d L Lxdt xx dt x x

δ δ δ

δ δ δ

δ δ

∂ ∂⎡ ⎤= +⎢ ⎥∂ ∂⎣ ⎦

⎡ ⎤ ⎡ ⎤∂ ∂ ∂⎡ ⎤ ⎛ ⎞ ⎛ ⎞= + −⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎢ ⎥∂ ∂ ∂⎣ ⎦ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

⎡ ⎤ ⎡ ⎤∂ ∂ ∂⎛ ⎞ ⎛ ⎞= − +⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

=

∫ ∫

∫ty)

Hence,

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ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

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Necessary Conditions

* Condition (1) must be satisfied regardless of the end condition.* Part of second equation may already be satisfied by the problem of specification. i.e. the amount of extra information contai

Note :

ned by thisequation varies with the boundary conditions specified.

1. 0 (Euler-Lagrange Equation)

2. 0 (Transversality Condition)f

o

t

t

L d Lx dt x

L xx

δ

∂ ∂⎛ ⎞ ⎛ ⎞− =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

⎡ ∂ ⎤⎛ ⎞ =⎜ ⎟⎢ ⎥∂⎝ ⎠⎣ ⎦

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ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

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Example – 1Problem: ( ) ( ) ( )

12

0

Minimize with 0 2, 1 3J x x dt x x= + = =∫

Solution:

1) E-L Equation:

( )2L x x= +

( )2

1 210 1 2 0,2 4

L d L tx x x t c t cx dt x∂ ∂⎛ ⎞− = ⇒ − = = ⇒ = + +⎜ ⎟∂ ∂⎝ ⎠

2) Boundary condition: ( ) 20 2x c= =

( )

( )

1

1

2

11 2 341 314 4

3Hence, 24 4

x c

c

t tx t

= + + =

= − =

= + +0

Transversality condition is automatically satisfied, since

0fx xδ δ= =

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ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

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Example – 2Problem: ( ) ( )

12

0

Minimize with 0 2,J x x dt x= + =∫

Solution:

1) E-L Equation:

( )2L x x= +

( )2

1 210 1 2 0,2 4

L d L tx x x t c t cx dt x∂ ∂⎛ ⎞− = ⇒ − = = ⇒ = + +⎜ ⎟∂ ∂⎝ ⎠

2) Boundary condition: ( ) 20 2x c= =

( )

( )

1 111

2

0, 0 0

12 0,2 2

Hence, 24 2

f f

f

f

f ft t

ft

t

L Lx xx x

tx c c

t tx t

δ δ

==

∂ ∂= ⇒ = ≠

∂ ∂

= + = ⇒ = −

= − +

( )1 : Freex

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ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

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Transversality Condition

0 0

0ff tt

t t

L Lx L x tx xδ δ⎡ ⎤∂ ∂⎡ ⎤ ⎧ ⎫+ − =⎨ ⎬⎢ ⎥⎢ ⎥∂ ∂⎣ ⎦ ⎩ ⎭⎣ ⎦

Special Cases:

( ) ( )0 01) Fixed End Points: , and , are fixed.

No additional information!f ft x t x

0

02) and are fixed (free initial and final states)

0f

f

t

t

t t

L xxδ∂⎡ ⎤ =⎢ ⎥∂⎣ ⎦

General condition:

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ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

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Transversality ConditionSpecial Cases:

0 03) , are fixed (free final time, free final state)

0

f f

f ft t

t x

L Lx L x tx x

δ δ∂ ∂⎧ ⎫+ − =⎨ ⎬∂ ∂⎩ ⎭

0 04) , and are fixed (free final time)

0f

f

t

t x x

LL xx∂⎧ ⎫− =⎨ ⎬∂⎩ ⎭

0 05) , and are fixed (free final state)

0

f

f

t

t x t

Lx∂

=∂

Page 26: Lecture 24

ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

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Transversality ConditionSpecial Cases:

( ) ( ) ( )0 06) , is fixed; , is constrained to lie on a given curve

0. However,

Hence, the transversality condition is

f ff

f

f f

f f f f f ft tt

ft

t x t x t

L L dx L x t x t tx x dt

L LL xx x

η

ηδ δ δ δ η δ

η

∂ ∂⎧ ⎫+ − = = =⎨ ⎬∂ ∂⎩ ⎭

∂ ∂+ −

∂ ∂( )0 0

f

f ft

t tδ δ⎡ ⎤⎧ ⎫⎢ ⎥ = ≠⎨ ⎬

⎩ ⎭⎢ ⎥⎣ ⎦

( ) 0ft

LL xx

η ∂⎧ ⎫+ − =⎨ ⎬∂⎩ ⎭Finally:

Page 27: Lecture 24

ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

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ExampleProblem: ( ) ( ) ( ) ( )

12

0

Minimize 1 with 0 0 and , lie on 5 15f fJ x dt x t x y t t= + = = − +∫Solution:

1) E-L Equation:

21L x= +

2

0

0 1 0

2

L d Lx dt x

d xdt x

ddt

∂ ∂⎛ ⎞− =⎜ ⎟∂ ∂⎝ ⎠∂⎛ ⎞− + =⎜ ⎟∂⎝ ⎠

2x

( )

2

2

22

2

212 10 0

11

2 1

x xx x xx

xx

x

+ −⎛ ⎞ += ⇒ =⎜ ⎟⎜ ⎟ ++⎝ ⎠

+( ) 22x x x− ( ) 0=

Page 28: Lecture 24

ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

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Example

1) E-L Equation: ( ) 1 20 0L d L x x t c t cx dt x∂ ∂⎛ ⎞− = ⇒ = ⇒ = +⎜ ⎟∂ ∂⎝ ⎠

2) Boundary condition:

( ) ( )

( )

( )

2 1

1 1

(1) 0 0

(2) 0

5 1 0 5 1 0 1/5

Hence, /5 To find , / 5 5 15 75/ 26

ft

f

f f f f

x c x t c t

LL y xx

x c c

x t tt t t t

= = ⇒ =

∂⎡ ⎤+ − =⎢ ⎥∂⎣ ⎦

− + = ⇒ − + = ⇒ =

=

= − + ⇒ =

Page 29: Lecture 24

Variational Problems in Multiple Dimensions: Without Constraints

Dr. Radhakant PadhiAsst. Professor

Dept. of Aerospace EngineeringIndian Institute of Science - Bangalore

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ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

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Multiple Dimension Problems without constraints

Problem: ( ) ( ) ( )0

Optimize , , by appropriate selection of .ft

t

J L X t X t t dt X t⎡ ⎤= ⎣ ⎦∫

Solution: ( )Make sure 0 for arbitrary J X tδ δ=

1) Euler – Lagrange (E-L) Equation

2) Transversality (Boundary) Condition

0L d LX dt X∂ ∂⎛ ⎞− =⎜ ⎟∂ ∂⎝ ⎠

Necessary Conditions:

[ ]1 2where TnX x x x

00

0f ft tT

T

tt

L LX L X tX X

δ δ⎡ ⎤ ⎡ ⎤⎧ ⎫∂ ∂⎛ ⎞ ⎛ ⎞+ − =⎢ ⎥ ⎨ ⎬⎢ ⎥⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠⎩ ⎭⎢ ⎥ ⎣ ⎦⎣ ⎦

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ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

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Variational Problems with Constraints

( )

( )

[ ] [ ]

0

1 2 1 2

Optimize : = , ,

Subject to : , , 0

where

,

ft

t

T Tn n

J L X X t dt

X X t

X x x x ϕ ϕ ϕ

Φ =

Φ

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ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

32

( ) ( )

( )0

1

*

Lagrange's Existence Theorem:

( ) : The above constrained optimizationproblem leads to the same solution as the followingunconstrained cost functional

= , , , ,

Let , ,

f

n

tT

t

t

J L X X t X X t dt

L X X t L

λ

λ

×∃

⎡ ⎤+ Φ⎣ ⎦

=

( ) ( ), , , ,TX X t X X tλ+ Φ

Variational Problems with Constraints

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ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

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* *

* *

Necessary Conditions of Optimality:(1) E-L Equations:

(a) 0 ( n equations)

(b)

L d LX dt X

L d Ldtλ λ

⎡ ⎤∂ ∂− =⎢ ⎥∂ ∂⎣ ⎦

⎡ ⎤∂ ∂− ⎢ ⎥∂ ∂⎣ ⎦

**

0 ( equations)

Note: 0 as there is no term in

n

L Lλλ

=

⎛ ⎞∂=⎜ ⎟∂⎝ ⎠

Variational Problems with Constraints

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ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

34

* **

*

(2) Transversality Conditions:

(a) 0

(b)

f f

oo

t tTT

tt

L LX L X tX X

L

δ δ

λ

⎡ ⎤ ⎡ ⎤⎧ ⎫⎛ ⎞ ⎛ ⎞∂ ∂⎪ ⎪⎢ ⎥ + − =⎢ ⎥⎨ ⎬⎜ ⎟ ⎜ ⎟∂ ∂⎢ ⎥ ⎪ ⎪⎢ ⎥⎝ ⎠ ⎝ ⎠⎩ ⎭⎣ ⎦⎣ ⎦

⎛ ⎞∂⎜ ⎟∂⎝ ⎠

**

f

o

tT

T

t

LLδλ λλ

⎡ ⎤ ⎛ ⎞∂⎢ ⎥ + − ⎜ ⎟⎢ ⎥ ∂⎝ ⎠⎣ ⎦0;

f

o

t

t

tδ⎡ ⎤⎧ ⎫⎪ ⎪⎢ ⎥ =⎨ ⎬⎢ ⎥⎪ ⎪⎩ ⎭⎣ ⎦

Variational Problems with Constraints

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ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

35

( )

( ) ( )

* *

*

0 0

* **

E-L Equations:

1) (a) 0

(b) , , 0 (same constraint equation)

2)Transversality Conditions: , fixed, , free

(a) f

f f

TT

ft

L d LX dt X

L X X t

t X t X

L LX L XX

λ

δ

⎛ ⎞ ⎛ ⎞∂ ∂− =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

⎛ ⎞∂= Φ =⎜ ⎟∂⎝ ⎠

⎛ ⎞∂ ∂+ −⎜ ⎟∂ ∂⎝ ⎠

*

*

0 ( equations)

(b) 0 However is free 0

so 0 ( 1 equation)

f

f

f

f

t

t f f f

t

t nX

L t t t

L

δ

δ δ

⎡ ⎤⎛ ⎞=⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦

= ⇒ ≠

=

Variational Problems with Constraints

Varaibles: 1 ( ) ( ) ( )

Boundary Conditions : 1f

n nX t

n n

λ+ +

+ +

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Constraint Equations

Nonholonomic constraints

Isoperimetric constraints

( ), , 0X X tΦ =

( )0

, ,

One way to get rid of Isoperimetric constraints is to convertthem into Nonholonomic constraints.

ft

t

q X X t dt k=∫

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37

Isoperimetric Constraints( )

0

1

1

1 1 0

1 1 0

1 0

1

Define: , ,

Then

( ) ( )

Choose one of ( ) or ( ) and fix the other

Let ( ) 0 ( )

f

n

t

nt

n f n

n f n

n

n f

x q X X t

x dt k

x t x t k

x t x t

x tx t k

+

+

+ +

+ +

+

+

=

=

− =

=

=

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38

( )1

1 0

1

The following additional non-holonomicconstraint is introduced:

, ,

with boundary conditions: ( ) 0 ( )

n

n

n f

x q X X t

x tx t

+

+

+

=

=

Summary :

The original problem is augmentedwith this information and solved.

k=

Isoperimetric Constraints

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39

References

T. F. Elbert, Estimation and Control Systems, Von Nostard Reinhold, 1984.

D. S. Naidu, Optimal Control Systems, CRC Press, 2002.

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